Review of conservative vector fields Recall that a vector field F is - - PowerPoint PPT Presentation

Review of conservative vector fields

Recall that a vector field F is conservative if there is a function f (the potential) such that F = ∇f . Let D = R2 ∖ {(0, 0)}, and let F be a vector field on D with continuous first order partial derivatives. Suppose that Py = Qx. Is F conservative? (a) Yes. (b) No. (c) Not enough information. (d) I don’t know.

Solution

There is not enough information. Consider the vector fields: F1(x, y) = ⟨ −2x (x2 + y2)2 , −2y (x2 + y2)2 ⟩ F2(x, y) = ⟨ −y x2 + y2 , x x2 + y2 ⟩ Both are defined over D = R2 ∖ (0, 0). Both satisfy Py = Qx. But F1 is conservative: it is the gradient of f (x, y) =

1 x2+y2 .

And F2 is not conservative: we saw earlier that if we integrate F2 around a circle containing the origin, we get 2π (and not 0).

Announcements

∙ Final exam is this Friday. Register for conflict by today,

Monday.

∙ Office hours/review session this week:

• Ordinary office hours Tuesday 11–11:50am.
• Extra office hours Wednesday evening (6–7pm—it’s fine with

me if you bring your dinner). AH 443 (Maybe also 5–6pm—sorry for lack of decision!)

• Extra office hours Thursday 12–1pm. AH 341
• Also office hours on Friday 9:30–10:30am. AH 341
• Come with questions (or you can listen to other people’s

questions). You can also post questions in advance on Piazza (there’s a folder called “questions-for-review-sessions” or something like that). ∙ TA help room—AH 147.

• Monday, Tuesday, Wednesday: 4–8pm.
• Thursday: 10–8pm. (Check back to confirm location.)
• Friday: no help room. (Maybe? I’m working on this. . . )

Other questions

Do you have severe allergies (such that you prefer people not bring those foods for their dinner to the review session)? (a) No severe allergies. (b) Severely allergic to peanuts. (c) Severely allergic to fish. (d) Severely allergic to something else, and I will email you about it today, so that you can make an announcement before the review sessions. (e) Severely allergic to stuff, but not planning on coming to the review session, so I don’t care if people bring it.

Other questions

Which chalk is best? (a) Option (a) (b) Option (b) (c) Option (c) (d) They’re all terrible, but I appreciate your effort anyway. I will now move to sit closer to the front of the room so I can see better.

Review of conservative vector fields: results in any dimension

Assumption: for today, all vector fields have continuous first order partial derivatives.

Theorem (Theorem A)

F is conservative ⇔ ∫︂

C

F · dr is path independent ⇔ ∫︂

C

F · dr = 0 for any closed path C. Method B F is conservative if we can find the potential f by hand. Recall: we solve for P = fx, Q = fy etc.

Results in R2

Suppose F = ⟨P, Q⟩, defined over D ⊂ R2.

Theorem (Theorem C2)

If F is conservative, then Py = Qx.

Theorem (Theorem D2)

If D is simply connected, and Py − Qx = 0, then F is conservative.

Recall the proof of Theorem D2

∙ By Theorem A, it’s enough to prove that

∫︁

C F · dr = 0 for any

closed path C in D.

∙ Step 1: We use Green’s theorem to show that

∫︁

C ′ F · dr = 0

for any simple closed path C ′ in D.

∙ Step 2: Then we show that any closed path C can be split

into a union of simple closed paths C1 ∪ C2 ∪ . . ..

∙ So

∫︂

C

F · dr = ∫︂

C1

F · dr + ∫︂

C2

F · dr + . . . = 0 + 0 + . . . by Step 1 = 0

Results in R3

Assume F = ⟨P, Q, R⟩ on D ⊂ R3.

Theorem (Theorem C3)

If F is conservative, then curl F = ⟨0, 0, 0⟩.

Theorem (Theorem D3)

If D = R3 and curl F = ⟨0, 0, 0⟩, then F is conservative.

Let’s prove Theorem D3

Compare to the proof of Theorem D2.

∙ By Theorem A, it’s enough to prove that

∫︁

C F · dr = 0 for any

closed path C in R3.

∙ Step 1: We use Stokes’ theorem to show that

∫︁

C ′ F · dr = 0

for any simple closed path C ′ in R3.

∙ Step 2: Then we show that any closed path C can be split

into a union of simple closed paths C1 ∪ C2 ∪ . . ..

∙ So

∫︂

C

F · dr = ∫︂

C1

F · dr + ∫︂

C2

F · dr + . . . = 0 + 0 + . . . by Step 1 = 0

Incompressible vector fields

Recall: We say that F is irrotational if curlF = ⟨0, 0, 0⟩. We say that F is incompressible if divF = 0.

Theorem (Theorem C3′)

If F = curl G, then div F = 0.

Theorem (Theorem D3′)

If F is defined on all of R3 and div F = 0, then F = curl G for some G.

Suppose you don’t know anything about D ⊂ R2, but I tell you that there is a vector field F = ⟨P, Q⟩ with Qx − Py = 0, but which is not conservative. What can you say about D? (a) It must be all of R2. (b) It must be simply connected. (c) It must not be simply connected. (d) It must be bounded. (e) I can’t say anything.

Solution

It must not be simply connected: If it were simply connected, then we could apply Theorem D2 (since Qx − Py = 0) and conclude that F is conservative, a contradiction.

The underlying math:

The more holes that D has, the more different vector fields F we can find which are not conservative but still satisfy Qx − Py = 0. So “counting” these vector fields tells us how many holes are in D. Going up one dimension, look at D ⊂ R3:

∙ We count vector fields which are irrotational (curlF = 0) but

not conservative. This tells us how many “one-dimensional holes” are in the solid D.

∙ We also count vector fields which are incompressible

(divF = 0) but not irrotational. This tells us how many “two-dimensional holes” are in the solid D. This is called studying the cohomology of the space D, and is a technique used in topology.